 # Let X_{1},\cdots,X_{n} be a random sample from f(x∣\theta)=\thetx^{\theta-1} for 0<x<1. dooporpplauttssg 2022-04-29 Answered

Let ${X}_{1},\cdots ,{X}_{n}$ be a random sample from $f\left(x\mid \theta \right)=\theta {x}^{\theta -1}$ for $0. Find a confidence interval for $\theta$ using as pivotal quantity a function of the maximum likelihood estimator for $\theta$.

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Step 1
Assuming you want a confidence interval for $\theta$, you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that

For the expected value we have the integral

And for the second moment

So the variance is
$Var\left(X\right)=\frac{\theta }{2+\theta }-{\left(\frac{\theta }{1+\theta }\right)}^{2}=$
$=\frac{\theta }{{\left(\theta +1\right)}^{2}\left(\theta +2\right)}.$
Now then, we know that the following quotient has a chi-squared distribution, so in this case

where ${S}^{2}=\frac{1}{n-1}\sum {\left({X}_{i}-\stackrel{―}{X}\right)}^{2}$. Finally we can work with the random variable C distributed as a chi squared. We want a confidence interval of $\left(1-\alpha \right)$. Then
$P\left({\chi }_{\left\{n-1\right\}}^{2}\left(\frac{\alpha }{2}\right)
denoting that ${\chi }_{\left\{n-1\right\}}^{2}\left(\beta \right)$ is the $\beta$ percentile of the ${\chi }^{2}$ which may be found in reference tables.
From here we work with the interval only substituting the percentiles with $\frac{1}{{D}_{t}}$ and $\frac{1}{{D}_{\pi }}$, lower and superior.
$\frac{1}{{D}_{l}}
${D}_{l}>\frac{\frac{\theta }{{\left(\theta +1\right)}^{2}\left(\theta +2\right)}}{\left(n-1\right){S}^{2}}>{D}_{s}$
${D}_{l}\left(\left(n-1\right){S}^{2}\right)>\frac{\theta }{{\left(\theta +1\right)}^{2}\left(\theta +2\right)}>{D}_{s}\left(\left(n-1\right){S}^{2}\right)$
$\left({D}_{l}\left(\left(n-1\right){S}^{2}\right)\right)>\frac{\theta }{{\left(\theta +1\right)}^{2}\left(\theta +2\right)}>\left({D}_{s}\left(\left(n-1\right){S}^{2}\right)\right)$
${\left(\stackrel{^}{\theta }+1\right)}^{2}\left(\stackrel{^}{\theta }+2\right)\left({D}_{l}\left(\left(n-1\right){S}^{2}\right)\right)>\theta >{\left(\stackrel{^}{\theta }+1\right)}^{2}\left(\stackrel{^}{\theta }+2\right)\left({D}_{s}\left(\left(n-1\right){S}^{2}\right)\right)$
where, taking the before definition and derivation, and ${S}^{2}=\frac{1}{n-1}\sum {\left({X}_{i}-\stackrel{―}{X}\right)}^{2}$. When we sent the $\theta$ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include $\theta$, then try it and tell us if it is simpler!